Explanation Given: • The rate (in liters per minute) at which water drains from a tank is recorded at half-minute intervals. • Time: 3 minutes. • Table: t (min) 0 0.5 1 1.5 2 2.5 3 r ( liters/min) 50 48 46 44 42 40 38 Key Concepts Used: Many applications are based on the following principle. The net change in a quantity s ( t ) is equal to the integral of its rate of change: s ( t 2 ) − s ( t 1 ) = ∫ s ' ( t ) ⋅ d t t 1 t 2 . Here s ( t 2 ) − s ( t 1 ) is the net change in s ( t ) over the time interval [ t 1 , t 2 ] and ∫ s ' ( t ) ⋅ d t t 1 t 2 is Integral of rate of change. Here in this problem, rate, r (in liters per minute) at which water drains from a tank is recorded at half-minute intervals and are tabulated in the given table. The data values are spaced at intervals Δ t = 0.5 minutes. We have to find N t h right-endpoint approximation ( R N ) and N t h left-endpoint approximation ( L N ). Then we have to take the average of N t h right-endpoint approximation ( R N ) and N t h left-endpoint approximation ( L N ) to find the total amount of water drained during the first 3 minutes. Formulas Used: N t h Right-endpoint approximation: R N = Δ x ∑ j = 0 N ( f ( a + Δ x ) + f ( a + 2 Δ x ) + ⋅ ⋅ ⋅ ⋅ ⋅ + f ( a + N Δ x ) ) . N t h Left-endpoint approximation: L N = Δ x ∑ j = 0 N − 1 ( f ( a ) + f ( a + Δ x ) + ⋅ ⋅ ⋅ ⋅ ⋅ + f ( a + ( N − 1 ) Δ x ) ) ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 5.6, Problem 18E

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Total amount of water drained during given time by using the average of left- and right-endpoint approximation

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Chapter 5.6, Problem 17E

Chapter 5.6, Problem 19E

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Chapter 5 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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Total amount of water drained during given time by using the average of left- and right-endpoint approximation

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Chapter 5 Solutions